Question

If $\sin B=\frac{1}{2}$, then $3\cos B-4{\cos }^{3}B$ is equal to

• A. $1$
• B. $\frac{3}{4}$
• C. $0$
• D. $2\frac{1}{2}$

Answer 1

The correct option is C $0$

$\sin B=\frac{1}{2}\cos B=\sqrt{1-{\sin }^{2}B}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

$\therefore B={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)={30}^o$

$\therefore 3\cos B-4{\cos }^{3}B=\cos 3B=\cos (3\times {30}^o)$

$=\cos {90}^o=0$